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Likurg_2 [28]
4 years ago
8

Technician A says that power is the rate that energy is stored. Technician B says that power refers to the rate that energy is t

ransformed into another kind of energy. Who is correct?
Select one:
a. Technician A
b. Technician B
c. Both technicians A and B
d. Neither technician A nor B
Physics
1 answer:
Alex777 [14]4 years ago
7 0

Answer:

C. Both technicians A and B

Explanation:

From the physical definition, power is defined as the rate of a body doing work. It is expressed as  

                                     P = w/t    watts

Where  

                       w - is the work done or the energy of the system in joules

                        t - time

The unit of power is represented in watts.

Whenever there is a rate of change of energy in the system, it accounts for the efficiency of the power of the system.

Hence, the statements of both technicians are correct.

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An ice skater accelerates backwards for 5.0 seconds to a final speed of 12 m/s. If the acceleration backwards was at a rate of 1
Lostsunrise [7]

Answer:

Vo = 4.5 [m/s]

Explanation:

In order to solve this problem, we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 12 [m/s]

Vo = initial velocity [m/s]

a = acceleration = 1.5 [m/s²]

t = time = 5 [s]

Now replacing:

12=v_{o}+1.5*5\\v_{o}=12- (7.5)\\v_{o}= 4.5[m/s]

4 0
3 years ago
What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh
Goshia [24]

Answer:

a) R_s = 0.092

b) R_p = 0.085

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

R_s = 0.092

b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

R_p = 0.085

3 0
3 years ago
Visible light is responsible for its color?
Gekata [30.6K]

Answer:

The colour of visible light depends on its wavelength. These wavelengths range from 700 nm at the red end of the spectrum to 400 nm at the violet end. Visible light waves are the only electromagnetic waves we can

Explanation:

5 0
3 years ago
Calculate the de Broglie wavelength of an electron traveling at 2.0 x 10^8 m/s (me = 9.1*10^-31 kg).
timofeeve [1]

Answer:

\lambda=3.64\times 10^{-12}\ m

Explanation:

We have,

Speed of an electron is 2\times 10^8\ m/s

It is required to find the De Broglie wavelength of electron. The formula for the De- Broglie wavelength is given by :

\lambda=\dfrac{h}{mv}

h is Planck's constant

m is mass of an electron

Plugging all the values we get :

\lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 2\times 10^{8}}\\\\\lambda=3.64\times 10^{-12}\ m

So, the De-Broglie wavelength of an electron is 3.64\times 10^{-12}\ m

4 0
3 years ago
1. Describe the vibration caused by primary waves.
mrs_skeptik [129]
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5 0
4 years ago
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